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Question
The equation of tangent to the curve y(1 + x2) = 2 – x, where it crosses x-axis is ______.
Options
x + 5y = 2
x – 5y = 2
5x – y = 2
5x + y = 2
Solution
The equation of tangent to the curve y(1 + x2) = 2 – x, where it crosses x-axis is x + 5y = 2.
Explanation:
Given that y(1 + x2) = 2 – x ...(i)
If it cuts x-axis, then y-coordinate is 0.
∴ 0(1 + x2) = 2 – x
⇒ x = 2
Put x = 2 in equation (i)
y(1 + 4) = 2 – 2
⇒ y(5) = 0
⇒ y = 0
Point of contact = (2, 0)
Differentiating equation (i) w.r.t. x, we have
`y xx 2x + (1 + x^2) "dy"/"dx"` = – 1
⇒ `2xy + (1 + x^2) "dy"/"dx"` = – 1
⇒ `(1 + x^2) "dy"/"dx"` = – 1 – 2xy
∴ `"dy"/"dx" = (-(1 + 2xy))/((1 + x^2))`
⇒ `"dy"/"dx"_(2, 0) = (-1)/((1 + 4)) = (-1)/5`
Equation of tangent is y – 0 = `- 1/5 (x - 2)`
⇒ 5y = – x + 2
⇒ x + 5y = 2
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